Calculate the volume of natural gas (methane) required to bring 3.9 L of water to a boil. Assume the initial temperature of the water is 39 C, and the methane is at 1 bar and 25 C. Assume that methane is an ideal gas, the specific heat capacity of water to be constant and equal to 4.184 J/g/K over this temperature range, and the density of water to be 1.0 g/cm3. The combustion of methane is given by
CH4 + 2 O2 → CO2 + 2 H2O ΔHcomb = -890.36 kJ/mol.
What amount of heat do you need to raise the temperature of H2O from 39 C to 100 C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
q = 3,900g x 4.184 x (100-39) = ? J energy needed.
Next, you know you obtain 890.36 kJ/mol CH4. How many moles CH4 do we need to supply ? J energy? That is
890,360 J/mol x y mol = ? J (from the water part above.) Solve for y mol
Last step, you have y mol CH4, which occupies 22.4L at STP. Convert y mol to L and convert to conditions in your problem if those conditions listed are not STP. I get confused between UK conditions and US conditions and new recommendations of IUPAC.
To calculate the volume of natural gas (methane) required to bring 3.9 L of water to a boil, we need to follow a few steps:
1. Convert the initial temperature of the water from Celsius to Kelvin.
- The initial temperature of the water is given as 39 C. To convert this to Kelvin, we add 273.15 to the Celsius value:
T1 = 39 C + 273.15 = 312.15 K
2. Calculate the change in enthalpy (∆H) required to boil the water.
- The enthalpy change (∆H) can be calculated using the specific heat capacity (C) and the mass of water (m) using the formula: ∆H = C * m * ∆T
Here, the specific heat capacity of water, C, is given as 4.184 J/g/K.
The mass of water, m, can be calculated using the density (d) and the volume (V) of water using the formula: m = d * V
The density of water, d, is given as 1.0 g/cm^3 and the volume of water, V, is given as 3.9 L.
- First, convert the volume of water from liters to grams since the density is given in g/cm^3:
V = 3.9 L * 1000 g/L = 3900 g
- Now calculate the mass of water:
m = 1.0 g/cm^3 * 3900 g = 3900 g
- Calculate the change in temperature (∆T) by subtracting the initial temperature from the boiling temperature of water (100 C or 373.15 K):
∆T = 373.15 K - 312.15 K = 61 K
- Calculate the enthalpy change (∆H):
∆H = 4.184 J/g/K * 3900 g * 61 K = 977,211.6 J
3. Convert the enthalpy change from J to kJ.
- Divide the enthalpy change (∆H) by 1000 to convert it from Joules to kilojoules:
∆H = 977,211.6 J / 1000 = 977.2116 kJ
4. Calculate the number of moles of methane (CH4) needed to produce the required enthalpy change (∆H).
- The balanced equation for the combustion of methane shows that 1 mole of methane releases 890.36 kJ of energy.
- Since the enthalpy change (∆H) required is -977.2116 kJ (negative because energy is released), we can use a proportion to calculate the number of moles of methane needed:
(1 mol CH4) / (890.36 kJ) = (x mol CH4) / (-977.2116 kJ)
- Solve for x (the number of moles of methane):
x = (1 mol CH4 * -977.2116 kJ) / 890.36 kJ = -1.09704 mol CH4
5. Calculate the volume of methane gas required using the ideal gas law.
- The ideal gas law, PV = nRT, relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T).
- Rearrange the equation to solve for the volume (V):
V = (nRT) / P
- The pressure of the methane gas is given as 1 bar and the temperature is given as 25 C, which needs to be converted to Kelvin:
T2 = 25 C + 273.15 = 298.15 K
- The gas constant (R) is 0.0821 L·atm/(mol·K).
- Plug in the values to calculate the volume of methane gas:
V = (-1.09704 mol CH4 * 0.0821 L·atm/(mol·K) * 298.15 K) / 1 bar = -28.582 L·atm/bar
Note: The negative sign indicates that the reaction is exothermic and releases energy. However, volume cannot be negative, so we can take the absolute value of the result: V = 28.582 L·atm/bar.
Therefore, the volume of natural gas (methane) required to bring 3.9 L of water to a boil is approximately 28.582 L·atm/bar.